Mixing
FTC in Integral Equation
$begingroup$
I am working on the following exercise (Source is unknown, unfortunately):
Find all continuous, non-negative functions $f:Bbb{R}toBbb{R}$ satisfying
$$int_0^xf(t)dt=(f(x))^alpha +C$$
for some constant $Cneq 0$ and $alpha >1$.
Here is my attempt:
Because of FTC, we know that $(f(x))^alpha$ is a differentiable function (thus continuous), and because the composition of two continuous is also continuous, $f$ is continuous. Applying FTC we have that
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^alpha+CBig]
\ f(x) & =alpha f(x)^{alpha-1}f'(x)
\ f(x)(alpha f(x)^{alpha-2}f'(x)-1) & = 0
end{align*}
so $f(x)=0$ or $f'(x)=alpha f(x)^{2-alpha}$.
Now, because this equation is in the form $y'+p(t)y=0$, we can probably do something with integrating factors and solve this D.E. However, I don't want to use any differential equation knowledge to solve this. How can I proceed?
calculus proof-writing
asked Jan 24 at 19:13
高田航高田航
1,360418
$endgroup$
add a comment |
$begingroup$
I am working on the following exercise (Source is unknown, unfortunately):
Find all continuous, non-negative functions $f:Bbb{R}toBbb{R}$ satisfying
$$int_0^xf(t)dt=(f(x))^alpha +C$$
for some constant $Cneq 0$ and $alpha >1$.
Here is my attempt:
Because of FTC, we know that $(f(x))^alpha$ is a differentiable function (thus continuous), and because the composition of two continuous is also continuous, $f$ is continuous. Applying FTC we have that
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^alpha+CBig]
\ f(x) & =alpha f(x)^{alpha-1}f'(x)
\ f(x)(alpha f(x)^{alpha-2}f'(x)-1) & = 0
end{align*}
so $f(x)=0$ or $f'(x)=alpha f(x)^{2-alpha}$.
Now, because this equation is in the form $y'+p(t)y=0$, we can probably do something with integrating factors and solve this D.E. However, I don't want to use any differential equation knowledge to solve this. How can I proceed?
calculus proof-writing
asked Jan 24 at 19:13
高田航高田航
1,360418
$endgroup$
$begingroup$
I don't understand the grammar in your penultimate sentence. Are you saying you don't want to use any differential equation knowledge to solve this? Your stated form for the equation is wrong anyway; I recommend the Ansatz $g=f^beta$, choosing $beta$ so $g^prime$ is $g$-independent.
$endgroup$
– J.G.
Jan 24 at 19:18
$begingroup$
But you already used some differential equation to solve: the very first step !
$endgroup$
– DonAntonio
Jan 24 at 19:18
add a comment |
$begingroup$
I am working on the following exercise (Source is unknown, unfortunately):
Find all continuous, non-negative functions $f:Bbb{R}toBbb{R}$ satisfying
$$int_0^xf(t)dt=(f(x))^alpha +C$$
for some constant $Cneq 0$ and $alpha >1$.
Here is my attempt:
Because of FTC, we know that $(f(x))^alpha$ is a differentiable function (thus continuous), and because the composition of two continuous is also continuous, $f$ is continuous. Applying FTC we have that
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^alpha+CBig]
\ f(x) & =alpha f(x)^{alpha-1}f'(x)
\ f(x)(alpha f(x)^{alpha-2}f'(x)-1) & = 0
end{align*}
so $f(x)=0$ or $f'(x)=alpha f(x)^{2-alpha}$.
Now, because this equation is in the form $y'+p(t)y=0$, we can probably do something with integrating factors and solve this D.E. However, I don't want to use any differential equation knowledge to solve this. How can I proceed?
calculus proof-writing
asked Jan 24 at 19:13
高田航高田航
1,360418
$endgroup$
I am working on the following exercise (Source is unknown, unfortunately):
Find all continuous, non-negative functions $f:Bbb{R}toBbb{R}$ satisfying
$$int_0^xf(t)dt=(f(x))^alpha +C$$
for some constant $Cneq 0$ and $alpha >1$.
Here is my attempt:
Because of FTC, we know that $(f(x))^alpha$ is a differentiable function (thus continuous), and because the composition of two continuous is also continuous, $f$ is continuous. Applying FTC we have that
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^alpha+CBig]
\ f(x) & =alpha f(x)^{alpha-1}f'(x)
\ f(x)(alpha f(x)^{alpha-2}f'(x)-1) & = 0
end{align*}
so $f(x)=0$ or $f'(x)=alpha f(x)^{2-alpha}$.
Now, because this equation is in the form $y'+p(t)y=0$, we can probably do something with integrating factors and solve this D.E. However, I don't want to use any differential equation knowledge to solve this. How can I proceed?
calculus proof-writing
calculus proof-writing
asked Jan 24 at 19:13
高田航高田航
1,360418
asked Jan 24 at 19:13
高田航高田航
1,360418
edited Jan 24 at 19:20
高田航
asked Jan 24 at 19:13
高田航高田航
1,360418
asked Jan 24 at 19:13
高田航高田航
1,360418
asked Jan 24 at 19:13
高田航高田航
1,360418
1,360418
$begingroup$
I don't understand the grammar in your penultimate sentence. Are you saying you don't want to use any differential equation knowledge to solve this? Your stated form for the equation is wrong anyway; I recommend the Ansatz $g=f^beta$, choosing $beta$ so $g^prime$ is $g$-independent.
$endgroup$
– J.G.
Jan 24 at 19:18
$begingroup$
But you already used some differential equation to solve: the very first step !
$endgroup$
– DonAntonio
Jan 24 at 19:18
add a comment |
$begingroup$
I don't understand the grammar in your penultimate sentence. Are you saying you don't want to use any differential equation knowledge to solve this? Your stated form for the equation is wrong anyway; I recommend the Ansatz $g=f^beta$, choosing $beta$ so $g^prime$ is $g$-independent.
$endgroup$
– J.G.
Jan 24 at 19:18
$begingroup$
But you already used some differential equation to solve: the very first step !
$endgroup$
– DonAntonio
Jan 24 at 19:18
$begingroup$
I don't understand the grammar in your penultimate sentence. Are you saying you don't want to use any differential equation knowledge to solve this? Your stated form for the equation is wrong anyway; I recommend the Ansatz $g=f^beta$, choosing $beta$ so $g^prime$ is $g$-independent.
$endgroup$
– J.G.
Jan 24 at 19:18
$begingroup$
I don't understand the grammar in your penultimate sentence. Are you saying you don't want to use any differential equation knowledge to solve this? Your stated form for the equation is wrong anyway; I recommend the Ansatz $g=f^beta$, choosing $beta$ so $g^prime$ is $g$-independent.
$endgroup$
– J.G.
Jan 24 at 19:18
$begingroup$
But you already used some differential equation to solve: the very first step !
$endgroup$
– DonAntonio
Jan 24 at 19:18
$begingroup$
But you already used some differential equation to solve: the very first step !
$endgroup$
– DonAntonio
Jan 24 at 19:18
add a comment |
1 Answer
1
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oldest
votes
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$frac{df}{dx}=alpha f^{2-alpha}$. By separating variables, we get $f^{alpha-2}df=alpha dx$. By integrating both sides, we have $frac{1}{alpha-1}f^{alpha-1}=alpha x+C$. Therefore, $f(x)=(alpha(alpha-1)x+C)^{frac{1}{alpha-1}}$ if $alphaneq1$. When $alpha=1$ then $frac{df}{f}=dx$. Therefore we have $ln f=x+C$. Hence $f(x)=Ce^{x}$.
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
If you evaluate the given relation at $x=0$, you can see that the given answer is probably not correct.
$endgroup$
– Rigel
Jan 24 at 19:53
add a comment |
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$begingroup$
$frac{df}{dx}=alpha f^{2-alpha}$. By separating variables, we get $f^{alpha-2}df=alpha dx$. By integrating both sides, we have $frac{1}{alpha-1}f^{alpha-1}=alpha x+C$. Therefore, $f(x)=(alpha(alpha-1)x+C)^{frac{1}{alpha-1}}$ if $alphaneq1$. When $alpha=1$ then $frac{df}{f}=dx$. Therefore we have $ln f=x+C$. Hence $f(x)=Ce^{x}$.
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
If you evaluate the given relation at $x=0$, you can see that the given answer is probably not correct.
$endgroup$
– Rigel
Jan 24 at 19:53
add a comment |
$begingroup$
$frac{df}{dx}=alpha f^{2-alpha}$. By separating variables, we get $f^{alpha-2}df=alpha dx$. By integrating both sides, we have $frac{1}{alpha-1}f^{alpha-1}=alpha x+C$. Therefore, $f(x)=(alpha(alpha-1)x+C)^{frac{1}{alpha-1}}$ if $alphaneq1$. When $alpha=1$ then $frac{df}{f}=dx$. Therefore we have $ln f=x+C$. Hence $f(x)=Ce^{x}$.
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
If you evaluate the given relation at $x=0$, you can see that the given answer is probably not correct.
$endgroup$
– Rigel
Jan 24 at 19:53
add a comment |
$begingroup$
$frac{df}{dx}=alpha f^{2-alpha}$. By separating variables, we get $f^{alpha-2}df=alpha dx$. By integrating both sides, we have $frac{1}{alpha-1}f^{alpha-1}=alpha x+C$. Therefore, $f(x)=(alpha(alpha-1)x+C)^{frac{1}{alpha-1}}$ if $alphaneq1$. When $alpha=1$ then $frac{df}{f}=dx$. Therefore we have $ln f=x+C$. Hence $f(x)=Ce^{x}$.
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$frac{df}{dx}=alpha f^{2-alpha}$. By separating variables, we get $f^{alpha-2}df=alpha dx$. By integrating both sides, we have $frac{1}{alpha-1}f^{alpha-1}=alpha x+C$. Therefore, $f(x)=(alpha(alpha-1)x+C)^{frac{1}{alpha-1}}$ if $alphaneq1$. When $alpha=1$ then $frac{df}{f}=dx$. Therefore we have $ln f=x+C$. Hence $f(x)=Ce^{x}$.
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
answered Jan 24 at 19:29
NotoriousJuanGNotoriousJuanG
843
843
$begingroup$
If you evaluate the given relation at $x=0$, you can see that the given answer is probably not correct.
$endgroup$
– Rigel
Jan 24 at 19:53
add a comment |
$begingroup$
If you evaluate the given relation at $x=0$, you can see that the given answer is probably not correct.
$endgroup$
– Rigel
Jan 24 at 19:53
$begingroup$
If you evaluate the given relation at $x=0$, you can see that the given answer is probably not correct.
$endgroup$
– Rigel
Jan 24 at 19:53
$begingroup$
If you evaluate the given relation at $x=0$, you can see that the given answer is probably not correct.
$endgroup$
– Rigel
Jan 24 at 19:53
add a comment |
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$begingroup$
I don't understand the grammar in your penultimate sentence. Are you saying you don't want to use any differential equation knowledge to solve this? Your stated form for the equation is wrong anyway; I recommend the Ansatz $g=f^beta$, choosing $beta$ so $g^prime$ is $g$-independent.
$endgroup$
– J.G.
Jan 24 at 19:18
$begingroup$
But you already used some differential equation to solve: the very first step !
$endgroup$
– DonAntonio
Jan 24 at 19:18